Math 1320 - Lab 2
Name:
uNiD:
Due: 09/12
Today we will consider a series of particle physics type problems. Answer the questions below and be sure
explain your reasoning behind your approach to these questions with appropriate diagrams and complete
sentences. In your writing, try to be complete enough so that someone who has completed Math 1310 would
understand what you are doing and why you are doing it, i.e. you don’t have to explain what an integral is,
but you do have to explain why you chose a particular form for an integral that you use.
1. (10 points) Bubble chambers are (were) used in experiments to detect charged high energy particles.
The chamber is filled with superheated liquid hydrogen and when charged particles enter, they leave
an ionization track which forms a line of bubbles which can be visualized. Kinks in the track represent
where a decay event occured (a high energy particle decayed into another particle). Assume a high
energy particle enters the chamber and undergoes one decay event before decaying further. It (and its
descendent) leaves a track that can be represented in parametric form by
(
0.5 cos(t),
0 ≤ t ≤ 3π
4
x(t) =
3π
0.5 cos(3π/4.) + 0.1 cos(t − 3π/4.),
4 ≤ t ≤ 6π
(
3π
0.5 sin(t),
0≤t≤ 4
y(t) =
3π
0.5 sin(3π/4.) + 0.1 sin(t − 3π/4.),
4 ≤ t ≤ 6π
(
t,
0 ≤ t ≤ 3π
4
z(t) =
3π
3π/4. + t/2.,
4 ≤ t ≤ 6π
If the particle and its descendent travel at 0.8c (c is the speed of light), how much time does the particle
survive for? What about the particle and its descendent? You can leave your answers in terms of c.
Solution: The particle’s distance has two parametric functions defining it; one equation representing the path before the decay, for 0 ≤ t ≤ 3π
4 , and the other representing the path after the decay,
for 3π
≤
t
≤
6π
4
The particle and its descendants travel at 0.8c. To find the time the particle survives we need
to compute the distance (equivalent to the arc-length) that the particle travels. Since we have a
piecewise representation for the parametric path, we need to compute two separate integrals. The
parameter, t, is the natural choice of variable to integrate over. This leads to the integral representing
the length of the particle being.
3 s
Z4 π L1 =
dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
0
3
=
Z4 πp
0.25 sin2 t + 0.25 cos2 t + 12 dt
0
3
=
Z4 π
√
1 + 0.25 dt = 2.63
0
Similarly, for the curve showing the path the particle’s descendant took we have
Z6π
L2 =
s
dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
3
4π
Z6π r
1
=
0.01 sin2 (t − 3π/4) + 0.01 cos2 (t − 3π/4) + dt = 8.41
4
3
4π
With these results, we see the total time the particle survived for is
descendant is 2.63+8.41
.
0.8c
2.63
0.8c ,
and the particle and its
2. (10 points) An experimental physicist is able to confine two charged particles in such a way that they
are always between 1 and 2 angstroms apart and measures the average potential energy between them
to be 720 (eV). They are considering using two model equations for the potential energy between the
particles, the Lennard-Jones potential model and the Born-Huggins-Mayer potential model. Equations
for each potenital (in eV) as a function of separation distance (in angstroms) is given below. Which
model does a better job at matching the experiment (if we assume all separation distances are equally
likely). Let σ = 2 angstroms.
VLJ (r) = 4 ∗ (0.5) ∗
σ 12
r
VBHM (r) = e8.75(σ−r) −
−
σ 6 r
50 10
− 8
r6
r
Solution: We need to compute the average potential energy for the two models presented. We will
then compare these averages to 720eV, whichever model is closest to 720eV is the best match for
the experiment.
PLJ
1
=
2−1
Z2
Z2 " 12 6 #
2
2
−
dr = 719.56
VLJ (r) dr = 2
r
r
1
1
PBHM
1
=
2−1
Z2
Z2
VBHM (r) dr =
1
e8.75(2−r) −
50 10
− 8 dr = 710.00
r6
r
1
These results show that the Lennard-Jones potential model does a better job matching experimental
values for this particular experiment.
3. (10 points) Consider the bond energy between salt ions Na+ and Cl− . Positive and negative charges
form an attractive force fa proportional to the inverse square of the distance r between the charges,
which can cause the sodium and chloride to combine in a solid crystal
fa (r) = −
A
,
r2
where A is a positive constant and the negative sign − indicates that the force fa pulls the two opposing
charges together. For large distances r, the force is weak, but nearby the force gets stronger, much like
the attractive force of a magnet. The two atoms also have a repulsive force that prevents the atoms from
fusing infinitesimally close to each other.
fb (r) =
B
,
rn
where B is a positive constant and the force goes with the inverse nth power of distance (the power is
typically anywhere between the 7th and 13th power, depending on the particular atoms involved).
The net force fnet on the two ions is
fnet (r) = fb (r) + fa (r)
The ions are in a stationary equilibrium when their position r∗ is such that the net force is zero: fnet = 0.
(a) Solve for r∗ .
(b) From the equilibrium point, compute the bonding energy of the two molecules by computing the
work involved to move the two ions from r∗ , to a position infinitely far apart. That is, compute the
work done against the force fnet (r) as we pull apart the ions by making r go to infinity to verify that
the work required has a form rC∗ where C is a positive constant that depends on n and A.
Solution: The equilibrium point is the value of r such that the net force is equal to zero. The r at
which the net force is zero is denoted as r∗ . So,
(a)
fnet (r) = fb (r) + fa (r) = 0
A
B
= n
r2
r
B
= rn−2
A
1
(n−2)
B
∗
r =
or otherwise
A
B = A(r∗ )n−2 .
(b) We need to compute the work done by taking infinitesimally small work and summing(integrating)
them all up together. Another way to look at this is that we are computing the work done by
taking the product of the net force over small distances dr and summing(integrating) them all
up. The integral representation is given by
W (r∗ ) =
Z
Z
∞
dW =
fnet (r)dr
r∗
Z
∞
=
Ar−2 − Br−n dr
r∗
h
Br−n+1 i∞
= − Ar−1 +
1 − n r∗
B(r∗ )−n+1
= A(r∗ )−1 −
1−n
∗ n−2 ∗ −n+1
A(r
)
(r )
= A(r∗ )−1 −
1−n
∗ −1
A(r
)
= A(r∗ )−1 −
1−n
1 C
A
= ∗
= ∗ 1−
r
n−1
r
Note that we used the result, B = A(r∗ )n−2 to eliminate B. Thus, we see that the bonding
energy has the form rC∗ which means that ionic bonds that have an equilibrium separation are
easier to break.